For a general topological space, these definitions are not equivalent.
The following counterexample is essentially Example 7.1.6 in Bogachev's Measure Theory. Let $X$ be a Vitali set of outer measure $1$ in $[0,1]$, such that $X$ is not Lebesgue measurable (and in particular does not have Lebesgue measure zero), but every Lebesgue measurable subset of $X$ does have Lebesgue measure zero. JDH explains how to construct such a Vitali set in this question. Equip $X$ with its subspace topology. (In fact, $X$ is a separable metric space, though it is not completely metrizable.) Then the Borel sets in $X$ are of the form $A \cap X$, where $A$ is Borel in $[0,1]$. For such $A \cap X$, define a measure $\mu$ by $\mu(A \cap X) = m(A)$, where $A$ is Borel in $[0,1]$ and $m$ is Lebesgue measure. It's simple to verify that $\mu$ is a countably additive Borel measure on $X$, with $\mu(X) = 1$. Furthermore, $\mu$ is regular in the sense of your second definition: for a Borel set $A \cap X$ as above, since $A$ is Borel in $[0,1]$ and $m$ is regular, for each $\epsilon$ there are sets $C,U \subset [0,1]$ which are respectively closed and open in $[0,1]$, $C \subset A \subset U$, and $m(U \setminus C) < \epsilon$. But then we have that $C \cap X, C \cap U$ are respectively closed and open in $X$, $C \cap X \subset A \cap X \subset U \cap X$, and $$\mu((U \cap X) \setminus (C \cap X)) = \mu((U \setminus C) \cap X) = m(U \setminus C) < \epsilon.$$
On the other hand, $\mu$ is not regular in the sense of your first definition. If $K \subset X$ is compact, then it is also compact in $[0,1]$ and in particular is Lebesgue measurable, so must have $m(K) = 0$. Then $\mu(K) = 0$ as well. In particular, we cannot approximate $X$ (or any other set of positive $\mu$ measure) from below by compact sets.
If your topological space is Polish, then these two definitions are equivalent, and in fact every finite Borel measure on such a space satisfies both. This is Theorem 7.1.7 in Bogachev.
Best Answer
No. Simply consider the Lebesgue measure $\lambda$ on $\mathbb{R}$ and the set $$B := (\mathbb{R} \backslash \mathbb{Q}) \cap [0,1].$$ It is widely known that $\lambda(B)=1$. On the other hand, we cannot find any open set $U \subseteq \mathbb{R}$ such that $U \subseteq B$. Hence,
$$\sup\{\lambda(U); U \, \text{open}, U \subseteq B\} = 0.$$